# Geometric Series

A sequence of numbers $$\left\{ {{a_n}} \right\}$$ is called a geometric sequence if the quotient of successive terms is a constant, called the common ratio. Thus $${\frac{{{a_{n + 1}}}}{{{a_n}}}} = q$$ or $${a_{n + 1}} = q{a_n}$$ for all terms of the sequence. It's supposed that $$q \ne 0$$ and $$q \ne 1.$$

For any geometric sequence:

${a_n} = {a_1}{q^{n - 1}}.$

A geometric series is the indicated sum of the terms of a geometric sequence. For a geometric series with $$q \ne 1,$$

${S_n} = {a_1} + {a_2} + \ldots + {a_n} = {a_1}\frac{{1 - {q^n}}}{{1 - q}},\;\; q \ne 1.$

We say that the geometric series converges if the limit $$\lim\limits_{n \to \infty } {S_n}$$ exists and is finite. Otherwise the series is said to diverge.

Let

$S = \sum\limits_{n = 0}^\infty {{a_n}} = {a_1}\sum\limits_{n = 0}^\infty {{q^n}}$

be a geometric series. Then the series converges to $$\frac{{{a_1}}}{{1 - q}}$$ if $$\left| q \right| \lt 1,$$ and the series diverges if $$\left| q \right| \gt 1.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the sum of the first $$8$$ terms of the geometric sequence

$3,6,12, \ldots$

### Example 2

Find the sum of the series

$1 - 0,37 + 0,37^2 - 0,37^3 + \ldots$

### Example 1.

Find the sum of the first $$8$$ terms of the geometric sequence

$3,6,12, \ldots$

Solution.

Here $${a_1} = 3$$ and $$q = 2.$$ For $$n = 8$$ we have

${S_8} = {a_1}\frac{{1 - {q^8}}}{{1 - q}} = 3 \cdot \frac{{1 - {2^8}}}{{1 - 2}} = 3 \cdot \frac{{1 - 256}}{{\left( { - 1} \right)}} = 765.$

### Example 2.

Find the sum of the series

$1 - 0,37 + 0,37^2 - 0,37^3 + \ldots$

Solution.

This is an infinite geometric series with ratio $$q = -0,37.$$ Hence, the series converges to

$S = \sum\limits_{n = 0}^\infty {{q^n}} = \frac{1}{{1 - \left( { - 0,37} \right)}} = \frac{1}{{1 + 0,37}} = \frac{1}{{1,37}} = \frac{{100}}{{137}}.$